NCERT Class 10 Maths Chapter 7 – Coordinate Geometry Exercise 7.2 NCERT Solutions
NCERT Class 10 Maths Chapter 7 – Coordinate Geometry Exercise 7.2 NCERT Solutions
The academic team at Neutronclasses has meticulously crafted NCERT solutions for Chapter 7 Exercise 7.2 of Class 10 Mathematics, which focuses on Coordinate Geometry. These solutions comprehensively cover all the exercises within this chapter. Presented below are the step-by-step solutions to all the questions found in the NCERT textbook for Chapter 7.
Exercise 7.2
1. Find the coordinates of the point which divides the join of (- 1, 7) and (4, – 3) in the ratio 2:3.
Answer: Let P(x, y) be the required point. Using the section formula
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Therefore the point is (1,3).
2. Find the coordinates of the points of trisection of the line segment joining (4, -1) and (-2, -3).
Answer:
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Let P (x1,y1) and Q (x2,y2) are the points of trisection of the line segment joining the given points i.e., AP = PQ = QB
Therefore, point P divides AB internally in the ratio 1:2.
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Therefore P(x1,y1) = (2, -5/3)
Point Q divides AB internally in the ratio 2:1.
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Q (x2 ,y2) = (0, -7/3)
3. To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1 m each. 100 flower pots have been placed at a distance of 1 m from each other along AD, as shown in the following figure. Niharika runs 1/4th the distance AD on the 2nd line and posts a green flag. Preet runs 1/5th the distance AD on the eighth line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flagexactly halfway between the line segment joining the two flags, where should she post her flag?
Answer:
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4. Find the ratio in which the line segment joining the points (-3, 10) and (6, – 8) is divided by (-1, 6).
Answer:
Let the ratio in which the line segment joining ( -3, 10) and (6, -8) is divided by point ( -1, 6) be k:1.
Therefore, -1 = 6k-3/k+1
-k – 1 = 6k -3
7k = 2
k = 2/7
Therefore, the required ratio is 2:7.
5. Find the ratio in which the line segment joining A (1, – 5) and B (- 4, 5) is divided by the x-axis. Also find the coordinates of the point of division.
Answer:
Let the ratio in which the line segment joining A (1, – 5) and B ( – 4, 5) is divided by x-axis be k:1.
Therefore, the coordinates of the point of division is (-4k+1/k+1, 5k-5/k+1).
We know that y-coordinate of any point on x-axis is 0.
∴ 5k-5/k+1 = 0
Therefore, x-axis divides it in the ratio 1:1.
To find the coordinates let’s substitute the value of k in equation(1)
Required point = [(- 4(1) + 1) / (1 + 1), (5(1) – 5) / (1 + 1)]
= [(- 4 + 1) / 2, (5 – 5) / 2]
= [- 3/2, 0]
6. If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.
Answer:
Let A,B,C and D be the points (1,2) (4,y), (x,6) and (3,5) respectively.
7. Find the coordinates of a point A, where AB is the diameter of circle whose centre is (2, – 3) and B is (1, 4).
Answer:
Let (x,y) be the coordinate of A.
Since AB is the diameter of the circle, the centre will be the mid-point of AB.
now, as centre is the mid-point of AB.
x-coordinate of centre = (2x+1)/2
y-coordinate of centre = (2y+4)/2
But given that centre of circle is (2,−3).
Therefore,
(2x+1)/2=2⇒x=3
(2y+4)/2=−3⇒y=−10
Thus the coordinate of A is (3,−10).
8. If A and B are (–2, –2) and (2, –4), respectively, find the coordinates of P such that AP = 3/7 AB and P lies on the line segment AB.
Answer:
As given the coordinates of A(−2,−2) and B(2,−4) and P is a point lies on AB.
And AP = 3/7 AB
∴BP = 4/7
Then, ratio of AP and PB = m1:m2 = 3:4
Let the coordinates of P be (x,y).
∴ x = (m1x2 + m2x1) / (m1 + m2)
⇒ x = (3 × 2 + 4 × (−2)) / (3 + 4) = (6 − 8) / 7 = −2 / 7
And y = (m1y2 + m2y1) / (m1 + m2)
⇒ y = ((3 × (−4) + 4 × (−2)) / (3 + 4) = (−12−8) / 7 = −20 / 7
∴ Coordinates of P = −2 / 7, −20 / 7
9. Find the coordinates of the points which divide the line segment joining A (- 2, 2) and B (2, 8) into four equal parts.
Answer:
From the figure, it can be observed that points X,Y,Z are dividing the line segment in a ratio 1:3,1:1,3:1 respectively.
Using Sectional Formula, we get,
Coordinates of X = ((1 × 2 + 3 × (−2)) / (1 + 3), (1 × 8 + 3 × 2) / (1 + 3))
= (−1, 7/2)
Coordinates of Y = (2 − 2) / 2, (2 + 8) / 2 = (0,5)
Coordinates of Z = ((3 × 2 + 1 × (−2)) / (1 + 3), (3 × 8 + 1 × 2) / (1 + 3)
= (1, 13/2)
10. Find the area of a rhombus if its vertices are (3, 0), (4, 5), (-1, 4) and (-2,-1) taken in order. [Hint: Area of a rhombus = 1/2(product of its diagonals)]
Answer:
Let (3, 0), (4, 5), ( – 1, 4) and ( – 2, – 1) are the vertices A, B, C, D of a rhombus ABCD.
Length of the diagonal AC=
Length of the diagonal BD=.png)
Area of rhombus ABCD = 1/2 X 4√2 X 6√2= 24 square units.
Therefore, the area of a rhombus if its vertices are (3, 0), (4, 5), (-1, 4) and (-2,-1) taken in order, is 24 square units.